a very occasional diary

nondescript
[Nikita Danilov]

2009-08-21

a trivial exercise.



Let's find a sum

\(\sum_{n=1}^\infty {1\over{n(n+2)}}\)

There is a well-know standard way, that I managed to recall eventually. Given that

\({1 \over n(n+2)} = {1 \over 2}\cdot \left({1\over n} - {1\over n+2}\right)\)

the sum can be re-written as

\(\sum_{n=1}^\infty {1\over{n(n+2)}} = \sum_{n=1}^\infty {1 \over 2}\left({1\over n} - {1\over n+2}\right) = {1\over 2}\left({1\over 1} - {1\over 3} + {1\over 2} - {1\over 4} + {1\over 3} - {1\over 5} + {1\over 4} - {1\over 6} \cdots\right)\)

with almost all terms canceling each other, leaving

\(\sum_{n=1}^\infty {1\over{n(n+2)}} = {1\over 2}\left(1 + {1\over 2}\right) = {3\over 4}\)

While this is easy to check, very little help is given on understanding how to arrive to the solution in the first place. Indeed, the first (and crucial) step is a rabbit pulled sans motif out of a conjurer hat. The solution, fortunately, can be found in a more systematic fashion, by a relatively generic method. Enter generating functions.

First, introduce a function

\(f(t) = \sum_{n=1}^\infty {t^{n + 1}\over n}\)

The series on the right converge absolutely when |t| < 1, so one can define

\(g(t) = \int f(t) dt = \int \sum_{n=1}^\infty {t^{n + 1}\over n} = \sum_{n=1}^\infty \int {t^{n + 1}\over n} = \sum_{n=1}^\infty {t^{n + 2}\over {n(n+2)}} + C\)

with the sum in question being

\(\sum_{n=1}^\infty {1\over{n(n+2)}} = g(1) - C = g(1) - g(0)\)

Definition of the g function follows immediately from the form of the original sum, and there is a limited set of operations (integration, differentiation, etc.) applicable to g to produce f.

The rest is more or less automatic. Note that

\(- ln(1 - t) = t + {t^2\over 2} + {t^3\over 3} + \cdots\)

so that

\(f(t) = t^2 + {t^3\over 2} + {t^4\over 3} + \cdots = - t \cdot ln(1-t)\)

therefore

\(g(t) = - \int t \cdot ln(1-t) dt = \cdots = {1\over 4} (1 - t)^2 - {1\over 2} (1 - t)^2 ln(1 - t) + (1 - t) ln(1 - t) + t + C\)

where the integral is standard. Now,

\(g(1) - g(0) = 1 - {1\over 4} = {3\over 4}\)

VoilĂ !

And just to check that things are not too far askew, a sub-exercise in a pointless programming:
scala> (1 to 10000).map(x => 1.0/(x*(x+2))).reduceLeft(_+_)
res0: Double = 0.749900014997506


PS: of course this post is an exercise in tex2img usage.
PPS: Ed. 2022: tex2img is gone, switch to mathjax.

1 comment:

  1. that's not pointless because of the xs

    ReplyDelete